P.s. using AC, one can ''construct'' an example taking the measure defined by a free ultrafilter on $X$. But... I would prefer to avoid the use of AC.

P.p.s. I could ask the same question also for $E(\mu)=\infty$, since the unique examples that I have in mind make use of the existence of an invariant measure on an amenable group, which follows by Hahn-Banach.

I have the feeling that the problem is deeper: it is difficult to prove the existence of a finitely additive non countably additive probability measure without using some version of AC. I am just asking if there is any particular case..

The first question is equivalent to the existence of a non-principal ultrafilter: a measure has entropy 0 if and only if every set has measure 0 or 1. This question was asked (and answered) at mathoverflow.net/questions/15872/non-principal-ultrafilters-on The second question is equivalent to the existence of a strictly finitely additive measure. Given any strictly finitely additive measure $\mu$, you can take its average with the countably additive measure $\nu(n)=C/(n\log n)^2$ to get a measure with infinite entropy.
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Anthony QuasOct 17 '11 at 13:54

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Indeed, it is consistent with ZF that every finitely additive measure on a sigma-algebra is countably additive.
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Gerald EdgarOct 17 '11 at 13:58

1 Answer
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Posted above as a comment, but re-posted here as an answer following suggestion of @Benoit Kloeckner so original q does not appear unanswered.

The first question is equivalent to the existence of a non-principal ultrafilter: a measure has entropy 0 if and only if every set has measure 0 or 1. This question was asked (and answered) at Non-principal ultrafilters on ω

The second question is equivalent to the existence of a strictly finitely additive measure. Given any strictly finitely additive measure $\mu$, you can take its average with the countably additive measure $\nu(n)=C/(n\log n)^2$ to get a measure with infinite entropy.